Trigonometric Equations

1. The equation is rearranged to give $\frac{\sin(x)}{\cos(x)}\equiv\tan(x)=\frac{1}{\sqrt{2}}\approx 0.707$. Inverting tan gives the default solution of $x=\tan^{-1}(0.707)=35.26^\circ$ and we can see from the graph that solutions repeat every 180 degrees:
   The solutions are $x=-324.74^\circ, -144.74^\circ, 35.26^\circ, 215.26^\circ$.
2. Similarly to the above, the default solution of the equation is $x+75^\circ=\tan^{-1}(3)=71.57^\circ$. There will be other solutions but we need to assess the intervals in which they will lie. Let $\theta=x+75^\circ$. If $0^\circ\leq x\leq 360^\circ$ then $75^\circ\leq \theta \leq 435^\circ$. The default value is outside this range and the solutions on the tan graph that lie in this interval are $\theta=251.57^\circ, 431.57^\circ$. It follows that the solutions to the original equation are $x=176.57^\circ$ and $x=356.57^\circ$.

It is very important to note here that if $x$ can be between $-180^\circ$ and $180^\circ$ then $2x$ can be between $-360^\circ$ and $360^\circ$. Firstly, as suggested above, we write the equation in the form $\cos(2x)=1$. Rather than transform the graph of $\cos(x)$, let $\theta=2x$ and look at the graph of $\cos(\theta)$. When looking at the graph of $\cos(\theta)$ to find values where cos takes the value 1, it is important to check all values of $\theta$ between $-360^\circ$ and $360^\circ$.

On this interval, $\cos(\theta)=1$ when $\theta=2x$ is equal to $-360^\circ$, $0^\circ$ and $360^\circ$. This gives three solutions for $x$ on the interval $-180^\circ\leq x\leq 180^\circ$:

$x=-180^\circ$, $x=0^\circ$ and $x=180^\circ$.

(quadratic trigonometric equations)

Don’t begin by dividing by $\sin(x)$ – this is because if you divide by anything you assume that it isn’t 0 and, of course, it might be. Instead, subtract $\sin(x)$ from both sides and factorise:

$\sin(x)(\sin(x)-1)=0$

A product that results in 0 means that either factor can be 0 and so we have $\sin(x)=0$ or $\sin(x)-1=0$, i.e. $\sin(x)=1$.

By examining the graph of $\sin(x)$ (see more on trig graphs) and using the inverse sine button on your calculator to find the multiple solutions, we can see that the solutions on the interval $0\leq x\leq 360^\circ$ are $0^\circ, 90^\circ, 180^\circ,$ and $360^\circ$.

Firstly, recognise that this is almost a quadratic. Replacing $\sin^2(3\theta)$ with $1-\cos^2(3\theta)$ and rearranging, we have a quadratic in cos:

$ 6(1-\cos^2(3\theta))=4-\cos(3\theta)$

$\Longrightarrow 6\cos^2(3\theta)-\cos(3\theta)-2=0$

Next,  solve the quadratic using factorising:

$(2\cos(3\theta)+1)(3\cos(3\theta)-2)=0$

or $\cos(3\theta)=-\frac{1}{2}$ and $\cos(3\theta)=\frac{2}{3}.$

Note that if $0<\theta< 180^\circ$, then $0< 3\theta<540^\circ$. Solutions on this interval can be seen in the graph below where we let $x=3\theta$:

If the x solutions are 48.19, 120, 240, 311.81, 408.19 and 480, then the solutions to the original equation are $\theta=16.06^\circ, 40^\circ, 80^\circ, 103.94^\circ, 136.06^\circ, 160^\circ$.